1995 AHSME Problems/Problem 30

Problem

A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is

$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$

Solution

Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.

Now consider the diagonal from $(0,0,0)$ to $(3,3,3)$ . The midpoint of this diagonal is at $\left(\frac 32,\frac 32,\frac 32\right)$ . The plane that passes through this point and is orthogonal to the diagonal has the equation $x+y+z=\frac 92$ .

The unit cube with opposite corners at $(x,y,z)$ and $(x+1,y+1,z+1)$ is intersected by this plane if and only if $x+y+z < \frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3$ . Therefore the cube is intersected by this plane if and only if $x+y+z\in\{2,3,4\}$ .

There are six cubes such that $x+y+z=2$ : permutations of $(1,1,0)$ and $(2,0,0)$ .
Symmetrically, there are six cubes such that $x+y+z=4$ .
Finally, there are seven cubes such that $x+y+z=3$ : permutations of $(2,1,0)$ and the central cube $(1,1,1)$ .

That gives a total of $\boxed{19}$ intersected cubes.

Note that there are only 8 cubes that are not intersected by our plane: 4 in each of the two opposite corners that were connected by the original diagonal.

1995 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Final Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1995 AHSME Problems/Problem 30

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